Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 13x}{x + 2} = \dfrac{2x - 50}{x + 2}$
Solution: Multiply both sides by $x + 2$ $ \dfrac{x^2 - 13x}{x + 2} (x + 2) = \dfrac{2x - 50}{x + 2} (x + 2)$ $ x^2 - 13x = 2x - 50$ Subtract $2x - 50$ from both sides: $ x^2 - 13x - (2x - 50) = 2x - 50 - (2x - 50)$ $ x^2 - 13x - 2x + 50 = 0$ $ x^2 - 15x + 50 = 0$ Factor the expression: $ (x - 5)(x - 10) = 0$ Therefore $x = 5$ or $x = 10$ The original expression is defined at $x = 5$ and $x = 10$, so there are no extraneous solutions.